How do you verify the identity $\frac{1}{1 - sin θ} + \frac{1}{1 + sin θ} = 2 {sec}^{2} θ$ $1/(1-sintheta)+1/(1+sintheta)=2sec^2theta$ ?

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How do you verify the identity $\frac{1}{1 - sin θ} + \frac{1}{1 + sin θ} = 2 {sec}^{2} θ$ $1/(1-sintheta)+1/(1+sintheta)=2sec^2theta$ ?

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Answer 1 · 2016-09-25T11:06:04

See below.

Explanation:

$\frac{1}{1 - sin θ} + \frac{1}{1 + sin θ} = 2 {sec}^{2} θ$ $1/(1-sintheta)+1/(1+sintheta)=2sec^2theta$
$\frac{1}{1 - sin θ} + \frac{1}{1 + sin θ} = \frac{(1 + sin θ) + (1 - sin θ)}{(1 - sin θ) (1 + sin θ)} = \frac{2}{1 - {sin}^{2} θ} = \frac{2}{{cos}^{2} θ} = 2 {sec}^{2} θ$ $1/(1-sintheta)+1/(1+sintheta)=((1+sintheta)+(1-sintheta))/((1-sintheta)(1+sintheta))=2/(1-sin^2theta) = 2/cos^2theta = 2 sec^2theta$

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How do you verify the identity $\frac{1}{1 - sin θ} + \frac{1}{1 + sin θ} = 2 {sec}^{2} θ$ $1/(1-sintheta)+1/(1+sintheta)=2sec^2theta$ ?

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Explanation:

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How do you verify the identity $\frac{1}{1 - sin θ} + \frac{1}{1 + sin θ} = 2 {sec}^{2} θ$ $1/(1-sintheta)+1/(1+sintheta)=2sec^2theta$ ?